Leila is playing a carnival game in which she is given 4

Question

Manhattan Prep Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A. 1/5^4 B. 1/5^3 C. 6/5^4 D. 13/5^4 E. 17/5^4 OA E

fskilnik@GMATH · Answer

AAPL wrote:Manhattan Prep Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A. 1/5^4 B. 1/5^3 C. 6/5^4 D. 13/5^4 E. 17/5^4 $$P\left( {{\rm{success}}\,\,{\rm{in}}\,\,{\rm{any}}\,\,{\rm{throw}}} \right) = {1 \over 5}$$ $$? = P\left( {{\rm{success}}\,\,{\rm{in}}\,{\rm{all}}\,\,4\,\,{\rm{throws}}} \right) + P\left( {{\rm{success}}\,\,{\rm{in}}\,{\rm{exactly}}\,\,3\,\,{\rm{throws}}} \right)$$ $$\left. \matrix{ P\left( {{\rm{success}}\,\,{\rm{in}}\,{\rm{all}}\,\,4\,\,{\rm{throws}}} \right)\,\,\,\mathop = \limits^{{\rm{independency}}} \,\,\,{\left( {{1 \over 5}} \right)^4} = {1 \over {{5^4}}} \hfill \cr P\left( {{\rm{success}}\,\,{\rm{in}}\,{\rm{exactly}}\,\,3\,\,{\rm{throws}}} \right)\,\,\,\mathop = \limits^{{\rm{independency}}} \,\,\,C\left( {4,3} \right) \cdot {\left( {{1 \over 5}} \right)^3} \cdot {4 \over 5}\,\,\, = \,\,\,{{4 \cdot 4} \over {{5^4}}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {{1 + 16} \over {{5^4}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)$$ We follow the notations and rationale taught in the GMATH method. Regards, Fabio.

Brent@GMATPrepNow · Answer

AAPL wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?

A. 1/5^4
B. 1/5^3
C. 6/5^4
D. 13/5^4
E. 17/5^4
OA E

Given: P(succeeds on 1 throw) = 1/5

P(succeeds at least 3 times) = P(succeeds 4 times OR succeeds 3 times)
= P(succeeds 4 times) + P(succeeds 3 times)

P(succeeds 4 times)
P(succeeds 4 times) = P(succeeds 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time)
= P(succeeds 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time)
= 1/5 x 1/5 x 1/5 x 1/5
= 1/5�

P(succeeds 3 times)
Let's examine one possible scenario in which Leila succeeds exactly 3 times:
P(FAILS the 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time)
= P(FAILS the 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time)
= 4/5 x 1/5 x 1/5 x 1/5
= 4/5�
Keep in mind that this is only ONE possible scenario in which Leila succeeds exactly 3 times (Leila fails the 1st time).
Leila can also FAIL the 2nd time, or the 3rd time or the 4th time.
Each of these probabilities will also equal 4/5³
So, P(succeeds 3 times) = 4/5� + 4/5� + 4/5� + 4/5�
= 16/5�

So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times) + P(succeeds 3 times)
= 1/5� + 16/5�
= 17/5�

Answer: E

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

swerve · Answer

P (that she succeeds) = 1/5
P(of success exactly 3 times) = 1/5*1/5*1/5*4/5 = 4/5^4. Now, these can occur in 4!/3!*1! = 4. hence, P(success 3 times)= 4*4/5^4
P (success all 4 times)= 1/5^4
Answer= 16/5^4+1/5^4 = 17/5^4.

Regards!

Scott@TargetTestPrep · Answer

AAPL wrote:Manhattan Prep Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A. 1/5^4 B. 1/5^3 C. 6/5^4 D. 13/5^4 E. 17/5^4 OA E We need to determine the probability that Leila succeeds on exactly 3 throws OR on all 4 throws. Scenario 1: succeeds on exactly 3 throws We can let Y denote a successful throw and N denote a non-successful throw: P(Y-Y-Y-N) = 1/5 x 1/5 x 1/5 x 4/5 = 4/(5^4) However, we must account for the order of Y-Y-Y-N. Using our formula for indistinguishable items, Y, Y, Y, and N can be arranged in 4!/3! = 4 ways. Thus, the probability of succeeding on exactly 3 throws (out of 4 attempts) is 4/(5^4) x 4 = 16/(5^4). Scenario 2: succeeds on all 4 attempts P(Y-Y-Y-Y) = 1/5 x 1/5 x 1/5 x 1/5 = 1/(5^4) Thus, the probability of succeeding on 3 throws out of 4 or 4 throws out of 4 is 16/(5^4) + 1/(5^4) = 17/5^4. Answer: E

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